Question

The flask is charged with 0.80 g magnesium turnings, THF (45 mL), 4-bromo-N,N-dimethylaniline (5.0 g), and a small crystal of iodine. The mixture is warmed gently to reflux (set the hot plate to 200 °C) and maintained there for 30 min, during which time the original dark reddish-brown color changes to the typical “dirty dishwater” shade of the Grignard reagent. Check with your TA to ensure this has indeed happened.

The flask is cooled to room temperature (ice-water bath). Be sure to keep the solution dry. After 15 min. diethyl carbonate (0.49 g) in 5 mL of THF is added in one portion. The mixture is warmed to reflux for an additional 5 min, and then cooled again to room temperature (ice-water bath). Aqueous hydrochloric acid (15 mL of a 10% solution) is added

Write the theoretical yield of your product, crystal violet

Answer 1

Answer – We are given, mass of Mg = 0.80 g ,

Mass of 4-bromo-N,N-dimethylaniline(C₈H₁₀BrN) = 5.0 g , mass of diethyl carbonate = 0.49 g. Aqueous hydrochloric acid = 15 mL of a 10% solution

From the given procedure reactions as follow –

Now we need to calculate the limiting reactant and for that purpose we need to calculate the moles of each reactant as follow –

We know,

Moles = given mass / molar mass

Moles of Mg = 0.80 g /

                      = 0.80 / 24.305 g.mol^-1

                      = 0.0329 moles

Moles of 4-bromo-N,N-dimethylaniline = 5.0 g / 200.075 g.mol^-1

                                                                 = 0.0249 moles

In the first reaction there is formed 4-(N,N-Dimethyl)aniline magnesium bromide and we need to calculate the how many moles of 4-(N,N-Dimethyl)aniline magnesium bromide(C₈H₁₀BrMgN) formed in the first reaction –

Moles of C₈H₁₀BrMgN from Mg

From the balanced reaction

1 moles of Mg = 1 moles of C₈H₁₀BrMgN

So, 0.0329 moles of Mg = ?

= 0.0329 moles of C₈H₁₀BrMgN

Moles of C₈H₁₀BrMgN from C₈H₁₀BrN

From the balanced reaction

1 moles of C₈H₁₀BrN = 1 moles of C₈H₁₀BrMgN

So, 0.0249 moles of Mg = ?

= 0.0249 moles of C₈H₁₀BrMgN

So, moles of C₈H₁₀BrMgN = 0.0249 moles , since the moles of C₈H₁₀BrMgN is lowest from the C₈H₁₀BrN, so this is got from the limiting reactant C₈H₁₀BrN.

Now moles of diethyl carbonate and moles of HCl

Mass of HCl – We are given the 15 mL of 10 % solution

We assume density of solution = 1.0 g/mL for aqueous phase

So, mass of solution = 15.0 mL * 1.0 g/mL

                                  = 15.30 g

So 10 % means

Mass percent = mass of solute / mass of solution * 100 %

Mass of HCl = 10 % * 15.0 g / 100 %

                     = 1.5 g

Moles of diethyl carbonate(C₅H₁₀O₃) = 0.49 g / 118.13 g.mol^-1

                                                              = 0.00415

Moles of HCl = 1.5 g / 36.461 g.mol^-1

                       = 0.0411 moles

Now moles of crystal violet(C₂₅N₃H₃₀Cl) from the 4-(N,N-Dimethyl)aniline magnesium bromide(C₈H₁₀BrMgN)

From the balanced reaction –

3 moles of C₈H₁₀BrMgN = 1 moles of C₂₅N₃H₃₀Cl

So, 0.0249 moles of C₈H₁₀BrMgN = ?

= 0.00833 moles of C₂₅N₃H₃₀Cl

Moles of C₂₅N₃H₃₀Cl from diethyl carbonate(C₅H₁₀O₃)

From the balanced reaction –

1 moles of C₅H₁₀O₃ = 1 moles of C₂₅N₃H₃₀Cl

So, 0.00415 moles of C₅H₁₀O₃= ?

= 0.00415 moles of C₂₅N₃H₃₀Cl

Moles of C₂₅N₃H₃₀Cl from HCl

From the balanced reaction –

1 moles of HCl = 1 moles of C₂₅N₃H₃₀Cl

So, 0.0411 moles of HCl= ?

= 0.0411 moles of C₂₅N₃H₃₀Cl

So moles of C₂₅N₃H₃₀Cl is lowest from the diethyl carbonate(C₅H₁₀O₃), so

Moles of C₂₅N₃H₃₀Cl = 0.00415 moles

Theoretical yield of crystal violet(C₂₅N₃H₃₀Cl) = 0.00415 moles *407.979 g/mol

                                                                         = 1.69 g